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Question-90331




Question Number 90331 by ajfour last updated on 22/Apr/20
Commented by ajfour last updated on 22/Apr/20
Express maximum of x, in terms  of a,h,l,μ,g.
$${Express}\:{maximum}\:{of}\:{x},\:{in}\:{terms} \\ $$$${of}\:{a},{h},{l},\mu,{g}. \\ $$
Answered by mr W last updated on 25/Apr/20
Commented by mr W last updated on 26/Apr/20
the stick is in equilibrium under the  acting of three forces, therefore  these three must be in the same plane  and they must meet at the same point.  φ=tan^(−1) μ  tan θ=(h/( (√(a^2 +x^2 ))))  α=φ  β=θ−φ  ((SM)/(sin ((π/2)−θ−φ)))=((AM)/(sin φ))  ⇒SM=((cos (θ+φ) l)/(2 sin φ))=(l/2)(((cos θ)/μ)−sin θ)  ((MB)/(sin β))=((SM)/(sin ((π/2)−θ+β)))  ⇒MB=((sin (θ−φ))/(cos φ))(((cos θ)/μ)−sin θ)×(l/2)  ⇒MB=(((cos θ−μ sin θ)(sin θ−μ cos θ))/μ)×(l/2)  AB=(h/(sin θ))  [1+(((cos θ−μ sin θ)(sin θ−μ cos θ))/μ)]×(l/2)=(h/(sin θ))  sin θ[1+(((cos θ−μ sin θ)(sin θ−μ cos θ))/μ)]=((2h)/l)  ⇒sin^2  θ cos θ=((2μh)/((1+μ^2 )l))  with λ=((μh)/((1+μ^2 )l))  ⇒cos^3  θ−cos θ+2λ=0  cos θ=((2(√3))/3) sin ((1/3)sin^(−1) 3(√3)λ+((2kπ)/3)) (k=0,1,2)  generally we get two suitable values  for θ. the smaller one leads to larger  value of x, therefore we take only  this solution:  let (1/δ)=((2(√3))/3) sin [(1/3)sin^(−1) ((3(√3)μh)/((1+μ^2 )l))+((2π)/3)]  tan^2  θ=(h^2 /(a^2 +x^2 ))=(1/(cos^2  θ))−1=δ^2 −1  ⇒x^2 =(h^2 /(δ^2 −1))−a^2   ⇒x=(√((h^2 /(δ^2 −1))−a^2 ))
$${the}\:{stick}\:{is}\:{in}\:{equilibrium}\:{under}\:{the} \\ $$$${acting}\:{of}\:{three}\:{forces},\:{therefore} \\ $$$${these}\:{three}\:{must}\:{be}\:{in}\:{the}\:{same}\:{plane} \\ $$$${and}\:{they}\:{must}\:{meet}\:{at}\:{the}\:{same}\:{point}. \\ $$$$\phi=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$$\mathrm{tan}\:\theta=\frac{{h}}{\:\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\alpha=\phi \\ $$$$\beta=\theta−\phi \\ $$$$\frac{{SM}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta−\phi\right)}=\frac{{AM}}{\mathrm{sin}\:\phi} \\ $$$$\Rightarrow{SM}=\frac{\mathrm{cos}\:\left(\theta+\phi\right)\:{l}}{\mathrm{2}\:\mathrm{sin}\:\phi}=\frac{{l}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right) \\ $$$$\frac{{MB}}{\mathrm{sin}\:\beta}=\frac{{SM}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta+\beta\right)} \\ $$$$\Rightarrow{MB}=\frac{\mathrm{sin}\:\left(\theta−\phi\right)}{\mathrm{cos}\:\phi}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right)×\frac{{l}}{\mathrm{2}} \\ $$$$\Rightarrow{MB}=\frac{\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right)}{\mu}×\frac{{l}}{\mathrm{2}} \\ $$$${AB}=\frac{{h}}{\mathrm{sin}\:\theta} \\ $$$$\left[\mathrm{1}+\frac{\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right)}{\mu}\right]×\frac{{l}}{\mathrm{2}}=\frac{{h}}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{sin}\:\theta\left[\mathrm{1}+\frac{\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right)}{\mu}\right]=\frac{\mathrm{2}{h}}{{l}} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta=\frac{\mathrm{2}\mu{h}}{\left(\mathrm{1}+\mu^{\mathrm{2}} \right){l}} \\ $$$${with}\:\lambda=\frac{\mu{h}}{\left(\mathrm{1}+\mu^{\mathrm{2}} \right){l}} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{3}} \:\theta−\mathrm{cos}\:\theta+\mathrm{2}\lambda=\mathrm{0} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \mathrm{3}\sqrt{\mathrm{3}}\lambda+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$${generally}\:{we}\:{get}\:{two}\:{suitable}\:{values} \\ $$$${for}\:\theta.\:{the}\:{smaller}\:{one}\:{leads}\:{to}\:{larger} \\ $$$${value}\:{of}\:{x},\:{therefore}\:{we}\:{take}\:{only} \\ $$$${this}\:{solution}: \\ $$$${let}\:\frac{\mathrm{1}}{\delta}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{sin}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{3}}\mu{h}}{\left(\mathrm{1}+\mu^{\mathrm{2}} \right){l}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right] \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta=\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\theta}−\mathrm{1}=\delta^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{{h}^{\mathrm{2}} }{\delta^{\mathrm{2}} −\mathrm{1}}−{a}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\sqrt{\frac{{h}^{\mathrm{2}} }{\delta^{\mathrm{2}} −\mathrm{1}}−{a}^{\mathrm{2}} } \\ $$
Commented by ajfour last updated on 26/Apr/20
great way sir!  but two things  a little typo of no consequence    cos^3 θ−cos^2 θ+2λ=0  And what if ground point of  rod, is not about to slide, but  just the point resting on the   edge is about to slide, then  x_(max)  shall be greater or smaller  than the already found result?
$${great}\:{way}\:{sir}! \\ $$$${but}\:{two}\:{things} \\ $$$${a}\:{little}\:{typo}\:{of}\:{no}\:{consequence} \\ $$$$\:\:\mathrm{cos}\:^{\mathrm{3}} \theta−\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{2}\lambda=\mathrm{0} \\ $$$${And}\:{what}\:{if}\:{ground}\:{point}\:{of} \\ $$$${rod},\:{is}\:{not}\:{about}\:{to}\:{slide},\:{but} \\ $$$${just}\:{the}\:{point}\:{resting}\:{on}\:{the}\: \\ $$$${edge}\:{is}\:{about}\:{to}\:{slide},\:{then} \\ $$$${x}_{{max}} \:{shall}\:{be}\:{greater}\:{or}\:{smaller} \\ $$$${than}\:{the}\:{already}\:{found}\:{result}? \\ $$
Commented by mr W last updated on 25/Apr/20
Commented by ajfour last updated on 26/Apr/20
thanks for this additional  diagram help, Sir.
$${thanks}\:{for}\:{this}\:{additional} \\ $$$${diagram}\:{help},\:{Sir}.\: \\ $$
Commented by mr W last updated on 26/Apr/20
if the point on groud is not about to  slide, then φ is smaller than now   (tan^(−1) μ), that means the stick is  steeper and x is smaller.
$${if}\:{the}\:{point}\:{on}\:{groud}\:{is}\:{not}\:{about}\:{to} \\ $$$${slide},\:{then}\:\phi\:{is}\:{smaller}\:{than}\:{now}\: \\ $$$$\left(\mathrm{tan}^{−\mathrm{1}} \mu\right),\:{that}\:{means}\:{the}\:{stick}\:{is} \\ $$$${steeper}\:{and}\:{x}\:{is}\:{smaller}. \\ $$
Commented by mr W last updated on 26/Apr/20
Commented by mr W last updated on 26/Apr/20
but to be honest, i am uncertain if my  solution is correct. let b=(√(a^2 +x^2 ))  my solution says the foot the stick  can stay at any point on the green  semicircle with radius b. but this  can not be true. at the position A′D′  the stick obtains no friction along  the edge, i think for the case the  solution is correct. but at any other  position the stick can slide along the  edge, this friction plays a role. but  this is not considered in my solution,  because i don′t know how to treat.
$${but}\:{to}\:{be}\:{honest},\:{i}\:{am}\:{uncertain}\:{if}\:{my} \\ $$$${solution}\:{is}\:{correct}.\:{let}\:{b}=\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} } \\ $$$${my}\:{solution}\:{says}\:{the}\:{foot}\:{the}\:{stick} \\ $$$${can}\:{stay}\:{at}\:{any}\:{point}\:{on}\:{the}\:{green} \\ $$$${semicircle}\:{with}\:{radius}\:{b}.\:{but}\:{this} \\ $$$${can}\:{not}\:{be}\:{true}.\:{at}\:{the}\:{position}\:{A}'{D}' \\ $$$${the}\:{stick}\:{obtains}\:{no}\:{friction}\:{along} \\ $$$${the}\:{edge},\:{i}\:{think}\:{for}\:{the}\:{case}\:{the} \\ $$$${solution}\:{is}\:{correct}.\:{but}\:{at}\:{any}\:{other} \\ $$$${position}\:{the}\:{stick}\:{can}\:{slide}\:{along}\:{the} \\ $$$${edge},\:{this}\:{friction}\:{plays}\:{a}\:{role}.\:{but} \\ $$$${this}\:{is}\:{not}\:{considered}\:{in}\:{my}\:{solution}, \\ $$$${because}\:{i}\:{don}'{t}\:{know}\:{how}\:{to}\:{treat}. \\ $$
Commented by mr W last updated on 26/Apr/20
i hope you can bring some light into  this point: how is the contact force  and friction force between the stick  and the edge if the strick is not at  the position A′D′.
$${i}\:{hope}\:{you}\:{can}\:{bring}\:{some}\:{light}\:{into} \\ $$$${this}\:{point}:\:{how}\:{is}\:{the}\:{contact}\:{force} \\ $$$${and}\:{friction}\:{force}\:{between}\:{the}\:{stick} \\ $$$${and}\:{the}\:{edge}\:{if}\:{the}\:{strick}\:{is}\:{not}\:{at} \\ $$$${the}\:{position}\:{A}'{D}'. \\ $$
Commented by ajfour last updated on 26/Apr/20
normal reaction ⊥ to stick,  friction along the edge, sideways  to the length of stick, and  along the length of stick as well.  why not consider the stick to be  hinged at the bottom (but allowed  to incline any way) and then  solve the problem, Sir?  i shall try once..
$${normal}\:{reaction}\:\bot\:{to}\:{stick}, \\ $$$${friction}\:{along}\:{the}\:{edge},\:{sideways} \\ $$$${to}\:{the}\:{length}\:{of}\:{stick},\:{and} \\ $$$${along}\:{the}\:{length}\:{of}\:{stick}\:{as}\:{well}. \\ $$$${why}\:{not}\:{consider}\:{the}\:{stick}\:{to}\:{be} \\ $$$${hinged}\:{at}\:{the}\:{bottom}\:\left({but}\:{allowed}\right. \\ $$$$\left.{to}\:{incline}\:{any}\:{way}\right)\:{and}\:{then} \\ $$$${solve}\:{the}\:{problem},\:{Sir}? \\ $$$${i}\:{shall}\:{try}\:{once}.. \\ $$
Commented by ajfour last updated on 26/Apr/20
i had happened to throw a fallen  matchstick 3-4 dsys back ,(when  i was thinking of creating a  new physics question)  randomly on the table, over   which were lying two books  larger below and smaller above,  the matchstick happened to  rest like in the question, thus  nature itself answered my wish.  (thought of sharing this even,    with you Sir, hidden   communications).
$${i}\:{had}\:{happened}\:{to}\:{throw}\:{a}\:{fallen} \\ $$$${matchstick}\:\mathrm{3}-\mathrm{4}\:{dsys}\:{back}\:,\left({when}\right. \\ $$$${i}\:{was}\:{thinking}\:{of}\:{creating}\:{a} \\ $$$$\left.{new}\:{physics}\:{question}\right) \\ $$$${randomly}\:{on}\:{the}\:{table},\:{over}\: \\ $$$${which}\:{were}\:{lying}\:{two}\:{books} \\ $$$${larger}\:{below}\:{and}\:{smaller}\:{above}, \\ $$$${the}\:{matchstick}\:{happened}\:{to} \\ $$$${rest}\:{like}\:{in}\:{the}\:{question},\:{thus} \\ $$$${nature}\:{itself}\:{answered}\:{my}\:{wish}. \\ $$$$\left({thought}\:{of}\:{sharing}\:{this}\:{even},\right. \\ $$$$\:\:{with}\:{you}\:{Sir},\:{hidden}\: \\ $$$$\left.{communications}\right). \\ $$
Commented by mr W last updated on 26/Apr/20
i made an experiment on our dinner  table and realised that my solution  is not correct. i′ll continue to think  about it.
$${i}\:{made}\:{an}\:{experiment}\:{on}\:{our}\:{dinner} \\ $$$${table}\:{and}\:{realised}\:{that}\:{my}\:{solution} \\ $$$${is}\:{not}\:{correct}.\:{i}'{ll}\:{continue}\:{to}\:{think} \\ $$$${about}\:{it}. \\ $$

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